Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
<em></em>
The compound is sodium chloride
<h3>
Answer:</h3>
5.71 × 10² nm
<h3>
Explanation:</h3>
The product of wavelength and frequency of a wave gives the speed of the wave.
Therefore;
Velocity of wave = Wavelength × Frequency
c = f ×λ
In our case;
Frequency = 5.25 × 10^14 Hz
Speed of light = 2.998 × 10^8m/s
But;
λ = c ÷ f
= 2.998 × 10^8m/s ÷ 5.25 × 10^14 Hz
= 5.71 × 10^-7 m
But; 1 M = 10^9 nm
Therefore;
wavelength = 5.71 × 10^-7 × 10^9
= 5.71 × 10² nm
The wavelength of light wave 5.71 × 10² nm
Remember that a cation will be smaller than its neutral atom, and an anion will be larger than its neutral atom. This would automatically eliminate answer choices A and D.
Also keep in mind that atomic radii decreases from left to right as you move along a periodic table. It also decreases from bottom up.
Atomic radii increases as you move from right to left and as you go from up to down.
As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.
The correct answer is B. Br
